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It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of: $$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$

I thought to proceed in this manner:

We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to convert everything to sum and product of roots form, but this way is too complicated!

Please suggest a simpler process.

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$$3a^2+2a+1=0 \to 3a^2+3a+1=a\\3b^2+2b+1=0\to 3b^2+3b+1=b\\a-1=3a(a+1)\\b-1=3b(b+1)$$ so $$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\\ \left(\dfrac{-3a(a+1)}{1+a}\right)^3+\left(\dfrac{-3b(b+1)}{1+b}\right)^3\\=-27(a^3+b^3)=-27(s^3-3ps)\\=-27((\frac{-2}{3})^3-3(\frac{2}{3}.\frac{1}{3})\\=+8-18\\=-10$$where $$s=a+b\\p=ab$$

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  • $\begingroup$ Thank you this is much simpler! $\endgroup$ – akhmeteni Jul 3 '17 at 11:50
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Plug $x=\frac{1-y}{1+y}$ in the given equation. We get: $$\frac{3 (1-y)^2}{(y+1)^2}+\frac{2 (1-y)}{y+1}+1=0$$ Expanding and collecting, we have: $$y^2-2 y+3=0$$ whose solutions are $$y_1=\frac{1-a}{1+a};\;y_2=\frac{1-b}{1+b}$$ We also know that sum of roots is $s=y_1+y_2=2$ and product is $p=y_1y_2=3$.

The sum of cubes can be written as follows $$y_1^3+y_2^3=\left(y_1+y_2\right)^3-3y_1y_2(y_1+y_2)=s^3-3ps=8-18=-10$$ so we have $$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=-10$$

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  • $\begingroup$ This is my way(+1) $\endgroup$ – lab bhattacharjee Jul 5 '17 at 13:06
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The answer is $-10$. Find the common denominator: $$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\frac{(1-ab-(a-b))^3+(1-ab+(a-b))^3}{(1+ab+(a+b))^3}=$$ $$\frac{2\cdot\left(\frac23\right)^3+2\cdot 3 \cdot\left(\frac23\right)\cdot (a-b)^2}{(\frac{2}{3})^3}=\frac{2\cdot\left(\frac23\right)^3+4\cdot ((a+b)^2-4ab)}{(\frac{2}{3})^3}=\frac{-10(\frac{2}{3})^3}{(\frac23)^3}=-10.$$

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  • $\begingroup$ Wolfram Alpha (wolframalpha.com/input/…) says that the answer is $-10$, as suggested by @khosrotash. Please check your calculations again. $\endgroup$ – Toby Mak Jul 3 '17 at 11:57
  • $\begingroup$ @Toby Mak, thank you. I fixed it. $\endgroup$ – farruhota Jul 3 '17 at 12:26
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From sum and product, we have:$$a+b+ab+ab=0$$ $$a(b+1)=-b(a+1)$$$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=\left(1-\dfrac{2a}{1+a}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3=\left(1-\dfrac{2ab}{b(1+a)}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3=\left(1+\dfrac{2b}{1+b}\right)^3+\left(1-\dfrac{2b}{1+b}\right)^3$$ Letting $x=\tfrac{2b}{b+1}$, we have $(1+x)^3+(1-x)^3=2+6x^2=2+6\left(\dfrac{2b}{1+b}\right)^2$.

As, $3b^2+2b+1=0$, so $(b+1)^2=-2b^2$, and finally $2+6\left(\dfrac{2b}{1+b}\right)^2=2+6\cdot(-2)=-10.$

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